| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on table with string above |
| Difficulty | Standard +0.3 This is a standard conical pendulum problem with two straightforward parts: (i) requires resolving forces vertically with given speed to find normal reaction, and (ii) requires recognizing that maximum speed occurs when normal reaction becomes zero. Both parts use routine circular motion equations (T cos 30° = mg, T sin 30° = mv²/r) with no novel insight required, making it slightly easier than average. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(0.6 \times 1.5^2/(0.2\cos30°) = T\cos30°\) | M1 | Uses N2L horizontally with component of tension |
| \(T = 9\) N | A1 | |
| \(R = 0.6g - 9\sin30°\) | M1 | Resolves vertically, 3 terms |
| \(R = 1.5\) N | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(T\sin30° = 0.6g\) | M1 | Resolves vertically, 2 terms |
| \(0.6v^2/(0.2\cos30°) = 12\cos30°\) | M1 | |
| \(v^2 = 3,\ v = 1.73\) | A1 [3] |
## Question 3:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $0.6 \times 1.5^2/(0.2\cos30°) = T\cos30°$ | M1 | Uses N2L horizontally with component of tension |
| $T = 9$ N | A1 | |
| $R = 0.6g - 9\sin30°$ | M1 | Resolves vertically, 3 terms |
| $R = 1.5$ N | A1 [4] | |
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $T\sin30° = 0.6g$ | M1 | Resolves vertically, 2 terms |
| $0.6v^2/(0.2\cos30°) = 12\cos30°$ | M1 | |
| $v^2 = 3,\ v = 1.73$ | A1 [3] | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{af7d1fc8-5552-48b8-a359-895b2b5d3d6c-2_279_905_1560_621}
One end of a light inextensible string of length 0.2 m is attached to a fixed point $A$ which is above a smooth horizontal surface. A particle $P$ of mass 0.6 kg is attached to the other end of the string. $P$ moves in a circle on the surface with constant speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, with the string taut and making an angle of $30 ^ { \circ }$ to the horizontal (see diagram).\\
(i) Given that $v = 1.5$, calculate the magnitude of the force that the surface exerts on $P$.\\
(ii) Given instead that $P$ moves with its greatest possible speed while remaining in contact with the surface, find $v$.
\hfill \mbox{\textit{CAIE M2 2010 Q3 [7]}}